There are two ways to solve a system of equations algebraically: the elimination method and the substitution method. Though the elimination method can be used at anytime, there are certain problems that lend them self to the use of the substitution method. Now that you have seen some examples on solving systems of equations with substitution, I can move on to some problem solving. The topic of this post is setting up and solving a system of two equations with the substitution method. The problem that follows is the classic part + part = whole relationship. The two parts are the cost of a slice of pizza and the cost of a soda. The whole is the total. Since there are two variables in the problem, there must be two equations. Read more to find out how to set up and solve a system of linear equations.

Example 1

At Renaldi’s Pizza, a soda and two slices of the pizza of the day costs $10.25. A soda and four slices of the pizza of the day costs $18.75. Find the cost of each item.

As you can see, the solution begins with the defining of variables: s for the price of one soda and p for the price of 1 slice of pizza. This is a fairly easy system of linear equations to set up because of the simple part + part = whole relationship. The two parts are the cost of the soda and the cost of the pizza. The whole is represented by the total cost. Thus, equations A and B are written based on the first two sentences of the problem.

Problem Solving - System of Equations - Pilarski

Lines 3 and 4 are still equations A and B, but it is an equivalent system. The variable term in each equation was subtracted from each side to arrive at the new equations A and B. The equation in line 5 is from applying the substitution property to replace the variable s with the 10.25 – 2p from equation A in line 3. Line 6 is from by subtracting 4p and 10.25 from both sides of line 5. Finally, line 7 is the result of applying the division property of equality to line 6. The result: 1 slice of pizza (p) costs $4.25.

This problem is not completely solved. The cost of one soda is still unknown. To find the cost of the soda, the cost of a slice of pizza must be substituted into any of the equations A or B. It appears the original equation A is used in the diagram on line 8. Line 9 shows the substitution of 4.25 into an original equation. The equation in line 10 is possible by simplifying 2 times 4.25 and the result in line 11 is from subtracting 8.50 from both sides of 10.

The problem finishes up with a brief sentence explaining the results. I try to embed a video on all of my posts. The video is the same problem, but if you are an auditory learner, it might help you more to hear me saying the things I write about in my posts. Until next time.

There are two ways to solve a system of equations algebraically: the elimination method and the substitution method. Though the elimination method can be used at anytime, there are certain problems that lend them self to the use of the substitution method. The topic of this post is solving a system of two equations with the substitution method. The problems covered in this post are basic, that is if you think algebra 2 is a basic course! By basic, I mean there is no problem solving. We are given the equations and told to solve them. Once solving the system is learned, then we can look at problem solving related to solving a system of equations.

Example 1 – Solve the system of equations using the substitution method.

As you can see in the figure below, I have the problem worked out. I have the given equations label A and B and I will refer to them as such as I solve the problem in paragraph format. This is one way to write mathematics. If your teachers are like me, then I am sure many of them spend time getting you to explain your answers. Here is what I think would be a viable way for you to write out how to solve a problem. You may notice the green numbers to the right of the problem. This is so you can follow along with needing to count the lines I am referring to in the problem. Let’s get started solving a system of two equations.

The Figure Above - Solving a System of Equations

In the problem to the right, lines 1 and 2 of the represent the given system of equations. I labeled the first equation A and the second equation B.

I could have solve equation A for x or y because each term had a coefficient of 1 and I could have solved equation B for y because it had a coefficient of 1. From my diagram, line 3 contains equation A, which I decided to solve for y and the equivalent equation A is on line 4.

Using the substitution property of equality I replace the y in equation B with the 10 – x (line 5) and the new equation B is contained in line 6. Simplifying the equation by combining the 2x and –x gives line 7. The subtraction property of equality allows us to isolate the x with a value of 5 (line 8).

Since I was able to find an x value, I will be able to find a y value. In line 9, I used equation A to solve for y. I substituted 5 in for y , line 10. Subtracting 5 from both sides of the equation in line 10 will give the result for y. It happens to be 5.

Thus, this system is independent and consistent and has one solution of (5,5).

The video below covers the problem above and additional problem. Coming soon, parts 2 and 3 of this lesson on solving systems of equations, I will tackle to problem solving situations.